a car traveling initially at +7.0 m/s accelerates uniformly at the rate of +0.80m/s2 for a distance of 245 m. what is its velocity at the end of the acceleration?
Just get out your good old
v = at
s = 1/2 at^2
formulas and work with them a bit.
We want to find out how long it will take to go 245m under acceleration.
245 = 7t + .4t^2
t = 17.5 sec
v = 7 + .8*17.5 = 21.0 m/s
To find the velocity of the car at the end of the acceleration, we can use the equation:
v = u + a * t
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity (u) = +7.0 m/s (positive because it is in the forward direction)
Acceleration (a) = +0.80 m/s^2 (positive because it is in the same direction as the initial velocity)
Distance (s) = 245 m
First, we need to find the time (t). We can rearrange the equation for acceleration:
a = (v - u) / t
Since the car is accelerating uniformly:
v = u + a * t
Given that the initial velocity (u) is +7.0 m/s, we need to find the time (t):
0.80 m/s^2 = (v - 7.0 m/s) / t
Multiplying both sides by t:
0.80 m/s^2 * t = v - 7.0 m/s
Now, we can substitute v = u + a * t:
0.80 m/s^2 * t = (7.0 m/s) + (0.80 m/s^2 * t)
Simplifying the equation:
0.80 m/s^2 * t - 0.8 t = 7.0 m/s
Now, we can solve for t:
0.8 t = 7.0 m/s
t = 7.0 m/s / 0.80 m/s^2
t = 8.75 seconds
Now that we have the time, we can substitute it back into the equation:
v = u + a * t
v = 7.0 m/s + (0.80 m/s^2 * 8.75 s)
v = 7.0 m/s + 7.0 m/s
v = 14.0 m/s
Therefore, the velocity of the car at the end of the acceleration is +14.0 m/s.
To find the velocity of the car at the end of the acceleration, you can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
In the given problem, the initial velocity (u) is +7.0 m/s, the acceleration (a) is +0.80 m/s^2, and the distance (s) is 245 m.
First, let's square the initial velocity:
u^2 = (7.0 m/s)^2 = 49.0 m^2/s^2
Next, substitute the given values into the formula:
v^2 = 49.0 m^2/s^2 + 2 * 0.80 m/s^2 * 245 m
Multiply the acceleration by the distance:
v^2 = 49.0 m^2/s^2 + 2 * 0.80 m/s^2 * 245 m
v^2 = 49.0 m^2/s^2 + 392 m^2/s^2
v^2 = 441.0 m^2/s^2
To get the final velocity, take the square root of both sides of the equation:
v = √(441.0 m^2/s^2)
v ≈ 21.0 m/s
Therefore, the velocity of the car at the end of the acceleration is approximately +21.0 m/s.